Let - 0- 4 and t1 - tan-tan- t2 - tan- t3 - -tan- then t4 - - then

The correct option is A t_1 lt; t_2 lt; t_1 lt; t_4 t_1=(tanθ)^tanθ t_2=(tanθ)^θ t_3=(θ)^tanθ t_4=(θ)^θ Let tanθ =α⇒θ =1_ / ...

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Standard X

Mathematics

Trigonometric Identities

Let θ∈ 0,π ...

Question

Let $θ \in (0, \frac{π}{4})$ and $t_{1} = {(tan θ)}^{tan θ}$ , $t_{2} = {(tan θ)}^{cot θ}$ , $t_{3} = {(cot θ)}^{tan θ}$ then $t_{4} = {(cot θ)}^{cot θ}$ , then

t_{1} < t_{2} < t_{1} < t_{4}

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t_{2} > t_{1} > t_{3} > t_{4}

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t_{3} > t_{1} > t_{2} > t_{4}

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t_{2} > t_{3} > t_{1} > t_{4}

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Solution

The correct option is A $t_{1} < t_{2} < t_{1} < t_{4}$
$t_{1} = (tan θ)^{tan θ}$
$t_{2} = (tan θ)^{cot θ}$
$t_{3} = (cot θ)^{tan θ}$
$t_{4} = (cot θ)^{cot θ}$
Let $tan θ = α \Rightarrow cot θ = 1_{/ α}$
$θ \in (0, π_{/ 4}) \Rightarrow α \in (0, 1)$
$t_{1} = (α)^{α}$
$t_{2} = (α)^{1 / α}$
$t_{3} = (1 / α)^{α}$
$t_{4} = (1 / α)^{1 / α}$
$α < 1$ but $1_{/ α} > 1$
implies $t_{3}$ & $t_{4}$ are greater
Since, $\frac{1}{α} > 1$ & $α < \frac{1}{α}$
Therefore, ${(\frac{1}{α})}^{α} < {(\frac{1}{α})}^{1 / α}$
$\Rightarrow t_{3} / t_{4}$ …. $(1)$
Since $α < 1_{/ α}$ , its clear that, anything raised to $(1 / α)$ will be smaller than $(α)$
$\Rightarrow t_{2} > t_{1}$ …… $(2)$
Considering $(1)$ & $(2)$
$t_{2} < t_{1} < t_{3} < t_{4}$ .

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